Proving the volume of the sphere

Factual Questions
1

There is a real question following this discursus, so please
bear with me until I get there.

Once, I worked out a fairly rigorous proof of the well-known
formula, (4/3)(pi)r**3 which gives the volume of the sphere. My approach was essentially as follows:

(1) We have a sphere with radius R. Temporarily disregard the lower hemisphere and just work on the half the sphere.

(2) Imagine slicing that hemisphere horizontally into (n) slices. The slices all have the same thickness, so the
thickness of each is R/n.

(3) Assume that if we have a sufficiently large (n), we can treat the slices not as true spherical slices, but as thin cylinders.

(4) Because the slices are of uniform thickness, we know exactly how far up the ith slice from the bottom of the hemisphere is–(R/n). By use of the Pythagorean Theorem,
we can therefore calculate the radius ® of the slice, enabling us to express the volume of the cylinder in terms of ® and (n).

(5) The approximate volume can then be expressed as a sum
of cylindrical slices. A little algebra reveals that as (n)
increases without bound, the resulting answer is (2/3)(pi)r**3.

(6) Add the bottom hemisphere back and double the above result to get (4/3)(pi)r**3.

Now my question. Are there any other approaches that work?
Someone once told me it was possible if you imagined the sphere to be composed of an infinite number of cones or pyramids, with their apexes touching the center of the sphere, but I don’t see how that could work.

2

The ancient Greeks didn’t believe in limits, so they didn’t use the in proofs (see my sig line). If memory serves, Euclid proved that he volume of phere is (4/3) pi radius cubed by proving that it was neither smaller nor larger than that value. I’ll have to rummage through my copy of the Elements and erify it.
Another odd proof is to note that the surface area is 4 pi radius squared. If you multiply that times an incemental radius delta r you have the volume of a sperical shelll. Now just integrate with respect to radius and you get (4/3) pi radius cubed.

Of course, I haven’t told you how to prove that the surface area is 4 pi radius squared. But do I have to do everything?

3

OK, I finally get to see one of these threads early enough that I can help with readibility.

If you choose, you can create superscripts and subscripts in VB by using {sup}text{/sup} and {sub}other text{/sub}, replacing the curly brackets with straight ones, just as if you were making italics or bold text.

Thus, [sup]text[/sup] and [sub]other text[/sub].

Carry on.

4

Well, it seems to me that a quick way to find the volume of a sphere using cones is:

(1) Find the surface area of a sphere. (4piR[sup]2[/sup])

(2) Assume that the sphere is approximated by n cones or pyrimids.

(3) Each cone has a volume of 1/3baseheight.

(4) As n-> infinity, height->R and base->(surface area/n)

(5) Volume = n1/2R*(4piR[sup]2[/sup])/n = (4/3)piR[sup]3[/sup]

(As I refresh, I see this is sorta what Cal said, but, since I already wrote it, I’ll go ahead and post.)

5

It begins to chime.

I see basically that if you can tessellate the surface of the sphere into a large number of polygons (Hexagons?), all
we have to do calculate one pyramid and add them up. And since we do believe in limits, from there the rest is easy.

Could hexagons be used? I mean, could one divide up the
surface of a sphere using them and not have any overlapping
or blank areas?

Yes you probably will have to explain how to prove the surface area formula, but not just yet. I want to poke around with it and see if I can come up with it. I won’t
say my mathematical knowledge is like a swiss cheese, unless
it’s more like the holes rather than the cheese.

6

Fill 'er up with water, then dump it into a measuring cup. :smiley:

7

Or, if you already have the formula for the volume of the sphere, the formula for the surface area is the derivative of the volume, with respect to r. This is because the increase in volume with a small increase in radius is equal to the surface area, times the increase in radius.

In other words, dV = A*dr.

In order to derive either formula from scratch, you’re going to end up doing an integral somewhere or another (even if it’s disguised as something else, like in javaman’s or Euclid’s derivations). What coordinate system you use for the integration is your choice: javaman used cylindrical, there, but it’s also common to use spherical coordinates. You could also use Cartesian (ack!) or some other system (double ack!) but there’s not much point to making your life hard that way, unless your teacher assigns it for homework or something.

8

javaman, the answer to your question is no. It’s impossible to tessellate a sphere with a finite number of hexagons. If you want each of the shapes in the tesselation to be regular polygons, and all the same, then there are only five possible configurations, one corresponding to each of the five Platonic Solids. However, it’s not necessary for the shapes to all be the same in order for the proof to function. A bunch of irregularly-shaped pyramids would do it. Of course, the volume of a pyramid, if I recall rightly, is proved through Integral Calculus, so if you’re looking for a way to do it withouth Integration (as I once did), you’re out of luck.

9

I believe Archimedes discovered the following derivation, and have read that he was so proud of it that a geometrical figure related to it was put on his tombstone.

The cross-sectional area of the hemisphere as a function of z coordinate is pi*(R[sup]2[/sup]-z[sup]2[/sup]). Now think of the solid formed by a cylinder with a cone removed from it. If the cylinder has height and radius equal to R, and the cone has its apex at z=0 and base with radius R at z=R, each cross-section is a circle of radius R with a hole of radius z. Since the hemisphere and the cylinder with removed cone have the same height and same cross-sectional area at each value of z, they have the volume. Volume of the cylinder minus cone is piR[sup]3[/sup]-pi/3R[sup]3[/sup]/3=2/3piR[sup]3[/sup], same as the hemisphere.

To see without calculus that the volume of a cone is one-third area of base times height, start with the dissection of a cube into 3 identical pyramids and use a similar identical-cross-sections argument.

10

I realize it looks like I’m maliciously violating a rule of
board etiquette, since this post will put my thread right
back on top. But seriously, is there a way for the originator of a thread to easily find and older thread when
it’s been aged out of the recent pages?

11

Best solution I’ve heard is to bookmark the URL for the thread itself… You can also just page through the forum listing, or you can search for your name and the topic, but those can be pretty slow.
And don’t worry about the etiquette; folks “bump” their threads all the time, and it’s usually not considered rude. The exception, I suppose, would be if you’re the only person keeping the thread alive, and continually asking, “C’mon, doesn’t anyone else want to answer?”

12

But how can you use “a bunch of irregularly-shaped pyramids”? In the end don’t you have to have to have a small enough number of different shapes, and know how many of them (for a given N total portions) are comprised in the
total known surface area?

I think I’ll stick to slicing…Call me Globeslicer

PS I agree the Southern Hemisphere gets all the good stars

13

In addition to summing up short, flat cylinders, cones, or spherical shells, you can sum over cylndrical shells. Imagine a bunch of cans without top or bottom, stacked one inside the other, with varying heights.

14

javaman, you’re right - I was misinterpreting you. It seems that you want actually to solve for the volume of a bunch of pyramids which approximate a sphere, not just sort of wave your hand and say that it works, which was my idea, which, I see, is similar to the one zut posted way back up at the top, although I think that our method would take a little more refinement, if you wanted to use it, which you don’t, so nevermind.

However, I still don’t think that your idea would quite work, and here’s why. If I understand you correctly, you want to tesselate a sphere with a certain number of different shapes, and then solve for the volume of each of the pyramids this defines. The problem is that for any number of shapes that you pick, you can only get so far. Let me try to explain.

With just one shape, the closest you can get to a sphere is an icosahedron. That’s what I was addressing earlier. If you used two shapes, I believe that the closest you can get is the so-called truncated icosahedron, which uses pentagons and hexagons. With three shapes, well, I don’t really know what you could get, but you could imagine it to be rather round.

So it’s all well and good, since you can find the volume of an icosahedron, or of a truncated icosahedron, by basic (if not pretty) AlgebroTrignonometroGeometric methods. But, the problem (I think) is that no matter how many surface shapes you use, there’s always some upper limit. So, with a finite number of shapes, you’ll only be able to approximate the volume of the sphere - never solve for it with a limit. For that, you’d need an infinite number of different bases for your pyramids. And, of course, if all you want is an approximation, then I think Garfield226 has the best answer. :slight_smile:

15

Dammit, I was gonna say that! But as a guy that barely passed high school algebra/trig, I can attest that it’s by far the easiest method :slight_smile:

16

First, remember that any pyramid or cone has the same formula for it’s volume: 1/3baseheight.

Therefore, if I had a pyramid that was 1" high and had a base that was 1 square inch, the total volume would be 1/3 cubic inches.

If I had three pyramids, the total volume would be:

V = (1/3*B[sub]1[/sub]H[sub]1[/sub]) + (1/3B[sub]2[/sub]H[sub]2[/sub]) + (1/3B[sub]3[/sub]*H[sub]3[/sub])

If the heights are all the same, this simplifies to V = 1/3*(B[sub]1[/sub] + B[sub]3[/sub] + B[sub]3[/sub])*H[sub]1[/sub]. So if I had three pyramids, all 1" high, and the areas of the bases, added together (B[sub]1[/sub] + B[sub]3[/sub] + B[sub]3[/sub]), totaled one square inch, the total volume would be 1/3 cubic inches.

In the same way, if I had seventeen pyramids, all 1" high, and the areas of the bases, added together, totaled one square inch, the total volume would be 1/3 cubic inches. It doesn’t matter what the shape of the base of any of the seventeen is, and it doesn’t matter if the bases are the same shape or size.

So, if you think about it, it doesn’t matter how many pyramids I have, as long as I know the total area of all the bases added together. To find the volume of the sphere, just assume there are an infinite number of pyramids (or near-infinite, if you prefer); you don’t know how many, but you do know that *the total area of the bases of the pyramids is 4piR[sup]2[/sup].

17

How do you prove that the volume of a cone is 1/3 the volume of its circumscribed cylinder.

(I’m not arguing that it isn’t–just wondering what the proof is.)

18

You can easily do it with a li’l bit of calculus: Chop up the cone into thin disks of thickness dx. The radius of any disk is [R*(x/h)], where R is the radius of the circumscribed cylinder, h is the height of the cone, and x is the distance of the disk from the tip of the cone. The volume of the disk is (pi)[R*(x/h)][sup]2[/sup]dx. Then adding up all the disks,

Volume = Integral(x from 0 to h)[(pi)[R*(x/h)][sup]2[/sup]dx] = pihR[sup]2[/sup]/3

I’m not quite sure how to do it without the calculus, but it’s probably important to realize that the area of these thin disks is a square function of the x-coordinate, so if you graph the area of the disks versus x, you get a parabola. The area under the parabola is analagous to the volume of the cone, and that area is (1/3) height*length.

19

This was already described in the earlier post about how Archimedes found the volume of the sphere without the use of calculus or limits.

An easy way to cover a sphere with facets for use as pyramid bases is to just define facet edges by lines of longitude and latitude. This defines trapezoidal facets (or triangular at the poles) in a way which is convenient for taking the limit as the number of facets goes to infinity, and is the same as integrating using spherical coordinates.

The method for creating a geodesic dome can be extended to limit of an infinite number of facets. The difficultly with this to is that the facet geometry is not easy to define mathematically. A messy infinite series is likely to be the result, and you would probably have to recognize an infinite series that equals pi to simplify this formula.

Another way to find volume of sphere without calculus is to use Pappus’ Centroid Theorem which dates from about 300 A.D. From this theorem, volume of a solid of revolution is found from the geometry of the plane shape that is revolved about an axis to make the solid: V=2pi(Area of 2D shape)(radial coordinate of centroid of the 2D shape). For the sphere the revolved 2D shape a semi-circle: area=piR[sup]2[/sup]/2; centroid coordinate=4R/(3pi). Finding the centroid is the hard part, but with calculus or limits it is not difficult.

20

Mjollnir, I did this before I tried the sphere, and as I remember, the algebra was harder for the cone. I’ve learned
some more since then, so maybe I should go back and try it
again.

As for the zut’s question, it was this issue that led me to
investigate this whole area. The CA Museum of Science and Industry used to have a wonderful math exhibit where they
explained that Archimedes had established the fact to which
zut alluded. And sure enough, just by looking at the formulas for the sphere, cone, and cylinder you can easily
verify that the volume of the first two equals the volume of
the cylinder, if said cylinder is superscribed around both the cone and sphere. But obviously this wasn’t satisfactory
because it didn’t explain how he came by the volumes in the
first place.

I wish someone had shown me stuff like this when they were trying to teach me math in school…being able to get
behind the formulas, so to speak, and prove their truth is
what make math interesting.